1. Given condition: reflected and refracted rays are perpendicular

When the reflected and refracted rays make an angle of $90^\circ$ with each other, the angle of incidence is the Brewster angle condition:

$i + r' = 90^\circ$

Here:

• angle of reflection $= r$
• angle of incidence $= i$
• by law of reflection,

$r=i$

So,

$r + r' = 90^\circ$

Thus,

$r' = 90^\circ - r$

2. Apply Snell's law

Let the refractive indices of denser and rarer media be $\mu_d$ and $\mu_r$ respectively.

Then,

$\mu_d \sin i = \mu_r \sin r'$

Since $i=r$,

$\mu_d \sin r = \mu_r \sin r'$

Using $r' = 90^\circ - r$,

$\sin r' = \cos r$

Hence,

$\mu_d \sin r = \mu_r \cos r$

So,

$\frac{\mu_r}{\mu_d} = \frac{\sin r}{\cos r} = \tan r$

3. Use definition of critical angle

For light going from denser to rarer medium, critical angle $C$ satisfies:

$\sin C = \frac{\mu_r}{\mu_d}$

From above,

$\sin C = \tan r$

Therefore,

$C = \sin^{-1}(\tan r)$

4. Check options

• A: $\sin^{-1}(\tan r)$ ✅
• B: $\sin^{-1}(\cot r)$ ❌
• C: $\sin^{-1}(\tan r')$ ❌
• D: $\tan^{-1}(\sin i)$ ❌

So the correct option is A.